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TRIGONOMETRIC FORMIJL}E
<br />IIB 71 • B
<br />.,'i C • a U . CL � 'LE'
<br />1 `
<br />A b CAA
<br />Right Triangle Oblique Triangles
<br />Solution .of Right. Triangles
<br />�__ 3 a For Angle A. sin ao> , cos � , tan= h , cot' b = a see h, cosec = a
<br />Given Required - , ,
<br />B ,c " tan A = = cot B, c = a2 +7 = a 1 +2
<br />{ a, a A, B, a sin A = d =cos B, b = %/—(c +a) (e—a}
<br />G v 02
<br />A, a B, b, c B = 90°—A, b = a cotA, o = sin A.
<br />•A, a B, a, c B =90°—A, a a tan A, e,= a
<br />cos A.
<br />A,oB, a, b B=90°—A,a=esin A,b=e.cosA,
<br />Solution of !Oblique, Triangles
<br />Given Required a sin B°— d sin C
<br />C b sin A , C = '180 (A + B}, c. = sin.A
<br />b'sin Aa sin C
<br />A,a,a B, 0,C sin B=. a ,C=180'�A+B)�c—•sin A
<br />a, b, C' A, B;"c A+B=180°— C, tan (A—)?)= (q—b) tan ? {A+B)
<br />a + a
<br />a sin C
<br />sin A
<br />J a+b+e jl5—
<br />a, b, c A, B, Cs 2 sin _A—
<br />0 a
<br />sin,' a o , C=180°—(A+B)
<br />a+b+d
<br />a, b, c Area - 2
<br />A, b, c Areaarea b 0 sin A
<br />= 2
<br />4, a' sin B sin C "
<br />A, B, C, a Area area = 2 sin A
<br />REDUCTION TO HORIZONTAL
<br />Horizontal distance= Slope distance multiplied.by the
<br />cosine of the vertical angle. Thus: slope distance =319.4 ft.
<br />9at'ce Vert. angle=5010'. From Table, Page IX. cos 5'101=
<br />e a� 9939. Horizontal distance=319.•1X,9959=315.09 ft.
<br />�loQ n1e Horizontal distance also=Slone distance minus slope
<br />yect. R distance times (1 --cosine of vertical angle). PJith the
<br />same figures as in the preceding example, the follow -
<br />Horizontal distance ing result is obtained. Gusine 5° I0I=.9955.1—.9969=.0041.
<br />319"4X,0041=1.31.319.4-1.31=313.09 rt.
<br />When the rise is known, the horizontal distance is approximately:—the slope dist-
<br />ance less the square of the rise divided by twice the slope distance. Thus: rise =llft.,
<br />slope di5tance=302.6 ft. Horizontal distance=302.6— 14 X 14 =232.6-0.32=902.28 ft.
<br />2 X 302.6
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